Q64 / 160 · TS EAMCET 2007 · Mathematics

If $\theta$ lies in the first quadrant and $5 \tan \theta=4$, then $\frac{5 \sin \theta-3 \cos \theta}{\sin \theta+2 \cos \theta}$ is equal to

  • Correct answer
    A
    $\frac{5}{14}$
  • B
    $\frac{3}{14}$
  • C
    $\frac{1}{14}$
  • D
    $0$

Solution

Given, $\theta$ lies in the first quadrant $\begin{array}{rlrl}\text { and } & 5 \tan \theta & =4 \\ \Rightarrow & \tan \theta & =\frac{4}{5} \\ & \therefore & \sin \theta=\frac{4}{\sqrt{41}}, \cos \theta & =\frac{5}{\sqrt{41}}\end{array}$ Now, $\begin{aligned} \frac{5 \sin \theta-3 \cos \theta}{\sin \theta+2 \cos \theta} & =\frac{5 \times \frac{4}{\sqrt{41}}-\frac{3 \times 5}{\sqrt{41}}}{\frac{4}{\sqrt{41}}+2 \times \frac{5}{\sqrt{41}}} \\ & =\frac{20-15}{4+10} \\ & =\frac{5}{14}\end{aligned}$

Hence, the correct answer is option: 'A'.

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